An experiment yields 3 mutually exclusive and exhaustive events A, B, C. If P (A) = 2P (B) = 3P (C), then P (A) is equal to

Options

A. `1/11`
B. `2/11`
C. `3/11`
D. `6/11`

Answer

To solve the problem, we need to find the probability of event A given the relationships between the probabilities of events A, B, and C.

### Step-by-Step Solution:

1. **Define the Probabilities**:
Let the probability of event C be \( P(C) = x \). According to the problem, we have:
\[ P(A) = 2P(B) = 3P(C) \]
Therefore, we can express \( P(B) \) and \( P(A) \) in terms of \( x \):
\[ P(B) = \frac{1}{2}P(A) \quad \text{and} \quad P(A) = 3x \]

2. **Express \( P(B) \)**:
Since \( P(A) = 3x \), we can find \( P(B) \):
\[ P(B) = \frac{1}{2}P(A) = \frac{1}{2}(3x) = \frac{3x}{2} \]

3. **Write the Total Probability**:
Since A, B, and C are mutually exclusive and exhaustive events, their probabilities must sum to 1:
\[ P(A) + P(B) + P(C) = 1 \]
Substituting the expressions we have:
\[ 3x + \frac{3x}{2} + x = 1 \]

4. **Combine the Terms**:
To combine the terms, we can convert them to a common denominator. The common denominator for 1, 2, and 1 is 2:
\[ 3x + \frac{3x}{2} + \frac{2x}{2} = 1 \]
This simplifies to:
\[ \frac{6x}{2} + \frac{3x}{2} + \frac{2x}{2} = 1 \]
\[ \frac{11x}{2} = 1 \]

5. **Solve for \( x \)**:
Now we can solve for \( x \):
\[ 11x = 2 \implies x = \frac{2}{11} \]

6. **Find \( P(A) \)**:
Now we can find \( P(A) \):
\[ P(A) = 3x = 3 \times \frac{2}{11} = \frac{6}{11} \]

### Final Answer:
Thus, the probability of event A is:
\[ P(A) = \frac{6}{11} \]